Conservation of Linear Momentum and Covariance

[Cave Johnson]

Homework Statement

Assume two masses m₁^' and m₂^' are moving in the positive x-direction with velocities v₁^' and v₂^' as measured by an observer in S^' before a collision. After the collision, the two masses stick together and move with velocity v^' in S^'. Show that if an observer in S^' finds momentum conserved, so does an observer in S.

Homework Equations

Galilean Transformation:
x^' = x - vt
y^' = y
z^' = z
t^' = t

Conservation of momentum in inelastic collisions:
m₁v₁ + m₂v₂ = (m₁ + m₂)v_f

Linear momentum:
p = mv

The Attempt at a Solution

I know that this will involve the use of this part of the GT:
x^' = x - vt

I am confused on how to incorporate the conservation of momentum equation(s) into this, however.

Any help would be appreciated.

 

[Doc Al]

How would you transform the velocities?

 

[Cave Johnson]

How would you transform the velocities?

I am not sure. I don't quite understand how to use these transformations for anything other than coordinates (like measuring lengths). I try to find examples in my textbook or online but they are all very confusing or blocked by a pay wall...

 

[Doc Al]

I don't quite understand how to use these transformations for anything other than coordinates (like measuring lengths).

Given the coordinate transformations, you can derive the velocity transformations by taking the derivative with respect to time. (It's easy!)

 

[Cave Johnson]

Given the coordinate transformations, you can derive the velocity transformations by taking the derivative with respect to time. (It's easy!)

Wouldn't that just leave us with -v ?

 

[Doc Al]

Wouldn't that just leave us with -v ?

Nope. Write the x-coordinate transform and take the derivative of each term.

 

[Cave Johnson]

Nope. Write the x-coordinate transform and take the derivative of each term.

Taking the derivative of x - vt with respect to time gives -v...
d/dt x = 0
d/dt -vt = -v

 

[Doc Al]

Taking the derivative of x - vt with respect to time gives -v...
d/dt x = 0
d/dt -vt = -v

Careful! The derivative of x with respect to t is not zero. It's dx/dt, which is a velocity measured in the S frame. Try it once more.